Question: The side of the base of a square prism is decreasing at a rate of $7$ kilometers per minute and the height of the prism is increasing at a rate of $10$ kilometers per minute. At a certain instant, the base's side is $4$ kilometers and the height is $9$ kilometers. What is the rate of change of the surface area of the prism at that instant (in square kilometers per minute)? Choose 1 answer: Choose 1 answer: (Choice A) A $-204$ (Choice B) B $148$ (Choice C) C $204$ (Choice D) D $-148$ The surface area of a square prism with base side $s$ and height $h$ is $2s^2+4sh$.
Solution: Setting up the math Let... $s(t)$ denote the base side of the prism at time $t$, $h(t)$ denote the height of the prism at time $t$, and $S(t)$ denote the surface area of the prism at time $t$. We are given that $s'(t)=-7$ and $h'(t)=10$ (notice that $s'$ is negative). We are also given that that $s(t_0)=4$ and $h(t_0)=9$ for a specific time $t_0$. We want to find $S'(t_0)$. Relating the measures The measures relate to each other through the formula for the surface area of a square prism: $S(t)=2[s(t)]^2+4s(t)h(t)$ We can differentiate both sides to find an expression for $S'(t)$ : $S'(t)=4[s(t)s'(t)+s'(t)h(t)+s(t)h'(t)]$ Using the information to solve Let's plug ${s(t_0)}={4}$, ${s'(t_0)}={-7}$, ${h(t_0)}={9}$, and $C{h'(t_0)}=C{10}$ into the expression for $S'(t_0)$ : $\begin{aligned} S'(t_0)&=4[{s(t_0)}{s'(t_0)}+{s'(t_0)}{h(t_0)}+{s(t_0)}C{h'(t_0)}] \\\\ &=4[({4})({-7})+({-7})({9})+({4})(C{10})] \\\\ &=-204 \end{aligned}$ In conclusion, the rate of change of the surface area of the prism at that instant is $-204$ square kilometers per minute. Since the rate of change is negative, we know that the surface area is decreasing.